Thi*_*dge 6 if-statement r mutated dplyr
我有以下数据集:
observation <- c(1:10)
pop.d.rank <- c(1:10)
cost.1 <- c(101:110)
cost.2 <- c(102:111)
cost.3 <- c(103:112)
all <- data.frame(observation,pop.d.rank,cost.1, cost.2, cost.3)
我想在三年内分配以下金额:
annual.investment <- 500
我可以使用以下脚本在第一年执行此操作:
library(dplyr)
all <- all %>%
mutate(capital_allocated.5G = diff(c(0, pmin(cumsum(cost), annual.investment)))) %>%
mutate(capital_percentage.5G = capital_allocated.5G / cost * 100) %>%
mutate(year = ifelse(capital_percentage.5G >= 50, "Year.1",0))
但是当我第二年尝试这样做时,考虑到前一年的投资,代码不起作用.这是我尝试在mutate循环中放置一个ifelse语句,以便它不会覆盖前一年分配的钱:
all <- all %>%
mutate(capital_allocated.5G = ifelse(year == 0, diff(c(0, pmin(cumsum(cost), annual.investment))), 0) %>%
mutate(capital_percentage.5G = capital_allocated.5G / cost * 100) %>%
mutate(year = ifelse(capital_percentage.5G >= 50, "Year.2",0))
我希望数据看起来如下所示,其中分配的金额首先发送到上一年未完成100%的任何行.
capital_allocated.5G <- c(101, 102, 103, 104, 105, 106, 107, 108, 109, 55)
capital_percentage.5G <- c(100, 100, 100, 100, 100, 100, 100, 100, 100, 50)
year <- c("Year.1", "Year.1","Year.1", "Year.1","Year.1", "Year.2", "Year.2","Year.2", "Year.2","Year.2")
example.output <- data.frame(observation,pop.d.rank,cost, capital_allocated.5G, capital_percentage.5G, year)
编辑:cost.1是第1年的成本变量,cost.2是第2年的变量,cost.3是第3年的成本变量
编辑:以前接受的答案的问题
我意识到这最终会为capital_percentage.5G变量分配超过100.我创建了一个可重现的例子.我认为这涉及一些成本随着时间的推移而降低并且一些成本随着时间的推移而增加的事实.
这背后的逻辑是,当投资在一年内完成时,5G移动网络的部署成本会有特定成本,这就是成本列与该时间点相关的成本.一旦这项投资在一年内完成,我希望该函数能够提供capital_percentage.5G 100%,然后在未来几年不再为其分配资金.
如何获得它以使百分比值达到100的限制,并且更多的资本分配不会在以后分配给它？
observation <- c(1:10)
pop.d.rank <- c(1:10)
cost.1 <- c(101:110)
cost.2 <- c(110:101)
cost.3 <- c(100:91)
all <- data.frame(observation,pop.d.rank,cost.1, cost.2, cost.3)
capital_allocated.5G <- rep(0,10) ## initialize to zero
capital_percentage.5G <- rep(0,10) ## initialize to zero
year <- rep(NA,10) ## initialize to NA
all <- data.frame(observation,pop.d.rank,cost.1, cost.2, cost.3, capital_allocated.5G,capital_percentage.5G,year)
alloc.invest <- function(df, ann.invest, y) {
df %>% mutate_(cost=paste0("cost.",y)) %>%
mutate(capital_percentage.5G = capital_allocated.5G / cost * 100,
year = ifelse(capital_percentage.5G < 50, NA, year),
not.yet.alloc = ifelse(capital_percentage.5G < 100,cost-capital_allocated.5G,0),
capital_allocated.5G = capital_allocated.5G + ifelse(capital_percentage.5G < 100,diff(c(0, pmin(cumsum(not.yet.alloc), ann.invest))), 0),
capital_percentage.5G = capital_allocated.5G / cost * 100,
year = ifelse(is.na(year) & capital_percentage.5G >= 50, paste0("Year.",y), year)) %>%
select(-cost,-not.yet.alloc)
}
annual.investment <- 500
all <- alloc.invest(all,annual.investment,1)
print(all)
all <- alloc.invest(all,annual.investment,2)
print(all)
all <- alloc.invest(all,annual.investment,3)
print(all)
在第3年,在这里的最终投资分配中,capital_percentage.5G突然飙升至110%.
对于每年可能减少和增加的不同成本,我们根本不需要检查capital_percentage.5G更新时是否超过100%not.yet.alloc并且capital_allocated.5G:
library(dplyr)
alloc.invest <- function(df, ann.invest, y) {
df %>% mutate_(cost=paste0("cost.",y)) %>%
mutate(capital_percentage.5G = capital_allocated.5G / cost * 100,
year = ifelse(capital_percentage.5G < 50, NA, year),
not.yet.alloc = cost-capital_allocated.5G,
capital_allocated.5G = capital_allocated.5G + diff(c(0, pmin(cumsum(not.yet.alloc), ann.invest))),
capital_percentage.5G = capital_allocated.5G / cost * 100,
year = ifelse(is.na(year) & capital_percentage.5G >= 50, paste0("Year.",y), year)) %>%
select(-cost,-not.yet.alloc)
}
使用新的成本数据:
observation <- c(1:10)
pop.d.rank <- c(1:10)
cost.1 <- c(101:110)
cost.2 <- c(110:101)
cost.3 <- c(100:91)
像以前一样使用初始值列进行扩充:
capital_allocated.5G <- rep(0,10) ## initialize to zero
capital_percentage.5G <- rep(0,10) ## initialize to zero
year <- rep(NA,10) ## initialize to NA
all <- data.frame(observation,pop.d.rank,cost.1, cost.2, cost.3, capital_allocated.5G,capital_percentage.5G,year)
1年级:
annual.investment <- 500
all <- alloc.invest(all,annual.investment,1)
print(all)
## observation pop.d.rank cost.1 cost.2 cost.3 capital_allocated.5G capital_percentage.5G year
##1 1 1 101 110 100 101 100.00000 Year.1
##2 2 2 102 109 99 102 100.00000 Year.1
##3 3 3 103 108 98 103 100.00000 Year.1
##4 4 4 104 107 97 104 100.00000 Year.1
##5 5 5 105 106 96 90 85.71429 Year.1
##6 6 6 106 105 95 0 0.00000 <NA>
##7 7 7 107 104 94 0 0.00000 <NA>
##8 8 8 108 103 93 0 0.00000 <NA>
##9 9 9 109 102 92 0 0.00000 <NA>
##10 10 10 110 101 91 0 0.00000 <NA>
2年级:
all <- alloc.invest(all,annual.investment,2)
print(all)
## observation pop.d.rank cost.1 cost.2 cost.3 capital_allocated.5G capital_percentage.5G year
##1 1 1 101 110 100 110 100.00000 Year.1
##2 2 2 102 109 99 109 100.00000 Year.1
##3 3 3 103 108 98 108 100.00000 Year.1
##4 4 4 104 107 97 107 100.00000 Year.1
##5 5 5 105 106 96 106 100.00000 Year.1
##6 6 6 106 105 95 105 100.00000 Year.2
##7 7 7 107 104 94 104 100.00000 Year.2
##8 8 8 108 103 93 103 100.00000 Year.2
##9 9 9 109 102 92 102 100.00000 Year.2
##10 10 10 110 101 91 46 45.54455 <NA>
3年级:
all <- alloc.invest(all,annual.investment,3)
print(all)
## observation pop.d.rank cost.1 cost.2 cost.3 capital_allocated.5G capital_percentage.5G year
##1 1 1 101 110 100 100 100 Year.1
##2 2 2 102 109 99 99 100 Year.1
##3 3 3 103 108 98 98 100 Year.1
##4 4 4 104 107 97 97 100 Year.1
##5 5 5 105 106 96 96 100 Year.1
##6 6 6 106 105 95 95 100 Year.2
##7 7 7 107 104 94 94 100 Year.2
##8 8 8 108 103 93 93 100 Year.2
##9 9 9 109 102 92 92 100 Year.2
##10 10 10 110 101 91 91 100 Year.3
您的代码的最初问题是ifelse只根据条件提供输出开关,而不是cost在TRUE分支中使用的输入ifelse.因此,cumsum(cost)计算cumsum全部cost而不仅仅是TRUE分支的一部分ifelse.为了解决这个问题,我们可以定义以下函数,然后可以依次为每年执行该函数.
library(dplyr)
alloc.invest <- function(df, ann.invest, y) {
df %>% mutate(not.yet.alloc = ifelse(capital_percentage.5G < 100,cost-capital_allocated.5G,0),
capital_allocated.5G = capital_allocated.5G + ifelse(capital_percentage.5G < 100,diff(c(0, pmin(cumsum(not.yet.alloc), ann.invest))), 0),
capital_percentage.5G = capital_allocated.5G / cost * 100,
year = ifelse(is.na(year) & capital_percentage.5G >= 50, paste0("Year.",y), year)) %>%
select(-not.yet.alloc)
}
注意:
not.yet.alloc,我们从中计算cumsum年份分配的结果.mutate陈述.is.na(year)在设置之前还需要检查year.否则,先前year已经标记的将被覆盖.要使用此功能,必须先用一些初始值增加的输入数据capital_allocated.5G,capital_percentage.5G以及year:
capital_allocated.5G <- rep(0,10) ## initialize to zero
capital_percentage.5G <- rep(0,10) ## initialize to zero
year <- rep(NA,10) ## initialize to NA
all <- data.frame(observation,pop.d.rank,cost,capital_allocated.5G,capital_percentage.5G,year)
然后是第1年:
annual.investment <- 500
all <- alloc.invest(all,annual.investment,1)
print(all)
## observation pop.d.rank cost capital_allocated.5G capital_percentage.5G year
##1 1 1 101 101 100.00000 Year.1
##2 2 2 102 102 100.00000 Year.1
##3 3 3 103 103 100.00000 Year.1
##4 4 4 104 104 100.00000 Year.1
##5 5 5 105 90 85.71429 Year.1
##6 6 6 106 0 0.00000 <NA>
##7 7 7 107 0 0.00000 <NA>
##8 8 8 108 0 0.00000 <NA>
##9 9 9 109 0 0.00000 <NA>
##10 10 10 110 0 0.00000 <NA>
而对于2年级:
all <- alloc.invest(all,annual.investment,2)
print(all)
## observation pop.d.rank cost capital_allocated.5G capital_percentage.5G year
##1 1 1 101 101 100 Year.1
##2 2 2 102 102 100 Year.1
##3 3 3 103 103 100 Year.1
##4 4 4 104 104 100 Year.1
##5 5 5 105 105 100 Year.1
##6 6 6 106 106 100 Year.2
##7 7 7 107 107 100 Year.2
##8 8 8 108 108 100 Year.2
##9 9 9 109 109 100 Year.2
##10 10 10 110 55 50 Year.2
如果每年的成本不同,那么函数需要首先重新调整列capital_percentage.5G和可能的year列:
library(dplyr)
alloc.invest <- function(df, ann.invest, y) {
df %>% mutate_(cost=paste0("cost.",y)) %>%
mutate(capital_percentage.5G = capital_allocated.5G / cost * 100,
year = ifelse(capital_percentage.5G < 50, NA, year),
not.yet.alloc = ifelse(capital_percentage.5G < 100,cost-capital_allocated.5G,0),
capital_allocated.5G = capital_allocated.5G + ifelse(capital_percentage.5G < 100,diff(c(0, pmin(cumsum(not.yet.alloc), ann.invest))), 0),
capital_percentage.5G = capital_allocated.5G / cost * 100,
year = ifelse(is.na(year) & capital_percentage.5G >= 50, paste0("Year.",y), year)) %>%
select(-cost,-not.yet.alloc)
}
需要注意的是创建另一个临时列cost使用mutate_仅仅是为了方便为代价列需要动态地选择基于输入y(否则,我们需要使用mutate_所有的计算,这将是有点混乱).
与用于初始值同样增加更新后的数据capital_allocated.5G,capital_percentage.5G和year,1年:
annual.investment <- 500
all <- alloc.invest(all,annual.investment,1)
print(all)
## observation pop.d.rank cost.1 cost.2 cost.3 capital_allocated.5G capital_percentage.5G year
##1 1 1 101 102 103 101 100.00000 Year.1
##2 2 2 102 103 104 102 100.00000 Year.1
##3 3 3 103 104 105 103 100.00000 Year.1
##4 4 4 104 105 106 104 100.00000 Year.1
##5 5 5 105 106 107 90 85.71429 Year.1
##6 6 6 106 107 108 0 0.00000 <NA>
##7 7 7 107 108 109 0 0.00000 <NA>
##8 8 8 108 109 110 0 0.00000 <NA>
##9 9 9 109 110 111 0 0.00000 <NA>
##10 10 10 110 111 112 0 0.00000 <NA>
第2年:请注意,上一个资产的50%分配少于分配,因此它year仍然是NA.
all <- alloc.invest(all,annual.investment,2)
print(all)
## observation pop.d.rank cost.1 cost.2 cost.3 capital_allocated.5G capital_percentage.5G year
##1 1 1 101 102 103 102 100.00000 Year.1
##2 2 2 102 103 104 103 100.00000 Year.1
##3 3 3 103 104 105 104 100.00000 Year.1
##4 4 4 104 105 106 105 100.00000 Year.1
##5 5 5 105 106 107 106 100.00000 Year.1
##6 6 6 106 107 108 107 100.00000 Year.2
##7 7 7 107 108 109 108 100.00000 Year.2
##8 8 8 108 109 110 109 100.00000 Year.2
##9 9 9 109 110 111 110 100.00000 Year.2
##10 10 10 110 111 112 46 41.44144 <NA>
3年级:
all <- alloc.invest(all,annual.investment,3)
print(all)
## observation pop.d.rank cost.1 cost.2 cost.3 capital_allocated.5G capital_percentage.5G year
##1 1 1 101 102 103 103 100 Year.1
##2 2 2 102 103 104 104 100 Year.1
##3 3 3 103 104 105 105 100 Year.1
##4 4 4 104 105 106 106 100 Year.1
##5 5 5 105 106 107 107 100 Year.1
##6 6 6 106 107 108 108 100 Year.2
##7 7 7 107 108 109 109 100 Year.2
##8 8 8 108 109 110 110 100 Year.2
##9 9 9 109 110 111 111 100 Year.2
##10 10 10 110 111 112 112 100 Year.3
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